Digestion of cellobiose in cows produces two glucose units which is then absorbed into the blood and distributed into the cells for energy production. One glucose unit is catabolized in the muscle cell while the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATP molecules are released? Show your detailed computation.

Digestion of cellobiose in cows produces two glucose units which is then absorbed into the blood and distributed into the cells for energy production. One glucose unit is catabolized in the muscle cell while the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATP molecules are released? Show your detailed computation.

Human Physiology: From Cells to Systems (MindTap Course List)

9th Edition

ISBN:9781285866932

Author:Lauralee Sherwood

Publisher:Lauralee Sherwood

Chapter2: Cell Physiology

Section: Chapter Questions

Problem 10RE: Using the answer code on the right, indicate which form of energy production is being described: 1....

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Digestion of cellobiose in cows produces two glucose units which is then absorbed into the bloodand distributed into the cells for energy production. One glucose unit is catabolized in the muscle cellwhile the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATPmolecules are released? Please show your detailed computation.
Digestion of cellobiose in cows produces two glucose units which is then absorbed into the bloodand distributed into the cells for energy production. One glucose unit is catabolized in the muscle cellwhile the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATPmolecules are released? Show your computation.
Digestion of cellobiose in cows produces two glucose units which is then absorbed into the bloodand distributed into the cells for energy production. One glucose unit is catabolized in the muscle cellwhile the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATPmolecules are released? Show your detailed answer / computation
Digestion of cellobiose in cows produces two glucose units which is then absorbed into the bloodand distributed into the cells for energy production. One glucose unit is catabolized in the muscle cellwhile the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATPmolecules are released? Show your detailed computation of the problem.
Digestion of cellobiose in cows produces two glucose units which is then absorbed into the bloodand distributed into the cells for energy production. One glucose unit is catabolized in the muscle cellwhile the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATP molecules are released? Show your detailed computation of the problem.
An animal cell, roughly cubical in shape with side length of 10 μm, uses 109 ATP molecules every minute. assume that the cell replaces this ATP by the oxidation of glucose according to the overall reaction 6O2 + C6H12O6 →6CO2 + 6H2O and that complete oxidation of each glucose molecule produces 30 ATP molecules. how much oxygen does the cell consume every minute? How long will it take before the cell has used up an amount of oxygen gas equal to its own volume?
Stearic acid is an 18-carbon fatty acid. If a single molecular of stearic acid is within the cytosol of the cells: Describe the process by which stearic acid would be metabolised, beginning the molecule in the cytosol of the cell, ending with the creation of ATP, and assuming the cell has sufficient oxygen for all reactions to take place Showing all working, calculate how many ATP molecules could be generated from a single molecule of stearic acid in the cytosol of the cell
The concentration of glucose in your circulatory system is maintained near 5.0 mM by the actions of the pancreatic hormones glucagon and insulin. Glucose is imported into cells by protein transporters that are highly specific for binding glucose. Inside the liver cells the imported glucose is rapidly phosphorylated to give glucose-6-phosphate (G-6-P). This is an ATP-dependent process that consumes 1 mol ATP per mol of glucose. Given the steady-state intracellular concentrations below, calculate the theoretical maximum concentration of G-6-P inside a liver cell at 37 °C, pH = 7.2 when the glucose concentration outside the cell (i.e., [glucoseloutside) is 5.0 mM: ATP = 4.7 mM; ADP = 0.15 mM; P, = 6.1 mM For: ATP + H,O ADP + P + H* AG" = -30.5 kJ/mol and G-6-P + H,0 -→ Glucose + P AG" = -13.8 kJ/mol The glucose phosphorylation reaction is ATP + glucosenside » ADP + glucose-6-phosphate + H+
Some reaction components are shown on the left. Match them to the reactions catalyzed by glutamine synthetase and glutaminase in the cell. (Multiple components on the left may match to the same option on the right. Some righthand options may have no corresponding component.) Glutamine ATP NAD+ H₂O Clear All Glutamine synthetase only Glutaminase only Both glutamine synthetase and glutaminase Neither glutamine synthetase nor glutaminase
The average cell, at rest, hydrolyzes 10,000, 000 ATP molecules per second. You are studying the stem cell population found in the intestinal Crypts. In the intestine there are s total 5 x 1011 cell number. The stem cells have a 85% higher metabolism than an average resting cell. ATP hydrolysis yield 7.4 kCal/ Mole ATP. (Avogadro’s number is 6.023 x 1023) 1) How many moles of ATP are hydrolyzed per second? A) 15.3 moles/ sec. B) 7.06 x 10-6 moles/ sec. C) 5.4 x 10-2 moles/ sec. D) 25 moles/ sec. E) 225 moles/hour F) none of these
The average cell, at rest, hydrolyzes 10,000, 000 ATP molecules per second. You are studying the stem cell population found in the intestinal Crypts. In the intestine there are s total 5 x 1011 cell number. The stem cells have a 85% higher metabolism than an average resting cell. ATP hydrolysis yield 7.4 kCal/ Mole ATP. (Avogadro’s number is 6.023 x 1023) 2) How many Kcal of energy are used per day by this population of cells? A) 4.514 kCal/day B) 1.14 x 10-6 kCal/day C) 1.14 x 10-4 kCal/day D) 25 kCal/day E) 225 kCal
The average cell, at rest, hydrolyzes 10,000, 000 ATP molecules per second. You are studying the stem cell population found in the intestinal Crypts. In the intestine there are s total 5 x 1011 cell number. The stem cells have a 85% higher metabolism than an average resting cell. ATP hydrolysis yield 7.4 kCal/ Mole ATP. (Avogadro’s number is 6.023 x 1023) Assuming 32 ATP per glucose oxidative metabolically, how many moles of glucose are consumed per day? A) 12,000 mole of glucose/ day B) 1.14 mole of glucose/ day C) 1.14 x 10-7 mole of glucose/ day D) 4.9 x 10-7 mole of glucose/ day E) 225 mole of glucose/ day F) 3.8 x 10-14 mole of glucose/ day Please explain how to do this!!!